Question

Kyle, a 95.0 kg95.0 kg football player, leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.430 kg0.430 kg ball precisely at the peak of his jump, when he is 0.386 meters0.386 meters off the ground. He hits the ground 0.0371 meters0.0371 meters away from where he leapt. If the ball was moving horizontally when it was caught, how fast was the ball traveling?

Answer 1

By using equation of motion

Acceleration a =g = 9.8 m/s^2

Initial velocity u =0

Time needed to free fall a distance of 0.386 m
0.386 = 4.9t²
t = 0.280 s

In that time, his horizontal displacement was 0.0371 m, so his horizontal velocity was:

Distance = speed × time
0.0371 = v × 0.280
v = 0.1325 m/s

He must have gotten that velocity from the inelastic collision with the ball, so

0.1325 = (95×0 + 0.43×U) / (95 + 0.43)

U = 29.4 m/s